Question

Find dxy if y=e^(1/x^2) + (1/e^x^2)

Answer 1

what is \(dxy\)?
Is that suppose to be \(\dfrac{dy}{dx}\)?

Answer 2

Dy (with a tiny x in the middle of D and y). Im assuming its implicit differentiation

Answer 3

just use the product rule since both x^2 and e^x are functions of x.

the product rule states that dy/dx is the derivative of the first function times the second plus the first function times the derivative of the second.

so dy/dx = dydx(x^2) * (e^x) + (x^2) * dy/dx(e^x)

dy/dx (x^2) is 2x
dy/dx (e^x) is e^x

so you get:

2x* e^x + x^2 * e^x

now you can factor out x*e^x from both terms to get

xe^x(2+x)

Answer 4

\[\huge y=e^{\left(1/x^2\right)}+\dfrac{1}{e^{x^2}}\]

Hmm I don't see product rule anywhere. Are you sure you read the problem correctly @jk3012 ?

I know these can be tricky to read when they're not formatted :C

Answer 5

It looks like we're actually going to be applying the chain rule several times :O
Let's clean up the exponents before differentiating.\[\huge y=e^{\left(x^{-2}\right)}+e^{-x^2}\]Understand how I changed those? I took the terms out of the denominator and applied a negative power to them.

Answer 6

ah ok, i had originally gotten the conversion to negative power wrong

Answer 7

Maybe it makes more sense to write it like this.\[\huge y=e^{(x^{-2})}+e^{-\left(x^2\right)}\]The negative is being applied to the exponent on the X in the first term, and to the exponent on the \(e\) in the second term.

Answer 8

Need to see some steps after that?

Answer 9

yes please...

Answer 10

would it be [e^(x^-2)](-2x^-3) + [e^-(x^2)](-2x)?

Answer 11

\[\huge y'=e^{(x^{-2})}\color{royalblue}{(x^{-2})'}+e^{-\left(x^2\right)}\color{royalblue}{(-x^2)'}\]

Yesss very good! :)

Answer 12

and thats it right?

Answer 13

is there a need to simplify even more?

Answer 14

It looks a little nicer if you pull the x's to the front, but yes that's all! :)

Answer 15

And pull the negatives to the fronts also *

Answer 16

ok will do... thanks so much :)